x = 3 t^3 + 1
dx/dt = 9 t^2
d^2x/dt^2 = 18 t
so at t = 0 both acceleration and velocity are 0
velocity is always greater than 0 so the particle never backs up. Therefore all we need is the position at t = 5 minus the position at t = 1
x(5) = 251
x(1) = 7
------------
difference = 244
A particle moves on the x –axis so that its position at any time is given by x(t) = 2t3 + 1.
a. Find the acceleration of the particle at t = 0.
b. Find the velocity of the particle when its acceleration is 0.
c. Find the total distance traveled by the particle from t = 0 to t = 5.
2 answers
thank you!!!!!!!!!!! I don't think any of my teachers would deticate some of their time for us:D
2 answers